Question

A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of 30^o with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground (g = 10 m/s², sin 30^o = $\frac{1}{2}$, $\cos 30^{o} = \frac{\sqrt{3}}{2}$)

• A. 8.66 m
• B. 5.20 m
• C. 4.33 m
• D. 2.60 m

Answer 1

Answer: (A)

8.66 m

Answer 2

Simply we have to calculate the range of projectile

$R = \frac{u^{2}\sin 2\theta}{g} = \frac{(10)^{2}\sin(2 \times 30{^\circ})}{10}$

$R = 5\sqrt{3} = 8.66meter$